Question: $\begin{cases} h(1)=9 \\\\ h(n)=h(n-1) \cdot \dfrac{1}{9} \end{cases}$ $h(3)=$
Solution: ${h(2)}={h(1)}\cdot \dfrac{1}{9}={9}\cdot \dfrac{1}{9}={1}$ ${h(3)}={h(2)}\cdot \dfrac{1}{9}={1}\cdot \dfrac{1}{9}={\dfrac{1}{9}}$ $h(3)=\dfrac{1}{9}$